Question

A car is traveling at 24.0 m/s when the driver suddenly applies the brakes, causing the car to slow down with constant acceleration. The car comes to a stop in a distance of 120 m. How fast was the car moving when it was 60.0 m past the point where the brakes were applied?

Answer 1

Answer:

17 m/s

Explanation:

Data

initial speed, Vi = 24 m/s

distance travelled, x = 120 m

final speed, Vf = 0 m/s

With help of the next Uniformly Accelerated Motion equation, we can compute car acceleration.

Vf² = Vi² + 2ax

a = Vf² - Vi²/(2x) 

a = 0² - 24²/(2*120) 

a = -2.4 m/s²

Now, the final speed is the unknown and the distance is 60 m:

Vf² = Vi² + 2ax

Vf = √(24² + 2*(-2.4)*60)

Vf = 17 m/s

Answer 2

Answer:

The answer to your question is : vf = 15.18 m/s

Explanation:

Data

vo = 24 m/s

d = 120 m

vf = ? when d = 60.0 m

Formula

              vf² = vo²  + 2ad

For d =100m

            a = (vf² - vo²) / 2d

            a = (0 -24²) / 2(100)

            a = -576/200

             a = 2.88 m/s²

Now, when d = 60

        vf² = (24)² - 2(2.88)(60)

        vf² = 576 - 345.6

        vf² = 230.4

        vf = 15.18 m/s